transfer function and corner frequency

Discussion in 'Homework Help' started by zdzislavv, Jun 14, 2009.

Jun 5, 2009
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2. Jony130 AAC Fanatic!

Feb 17, 2009
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The gain of a circuit is equal

$Au=\frac{1-j\omega R1(C1+C2)}{1-(j \omega R1(C1+C2)^2}*Km*\frac{R3}{R2+R3}$

3. The Electrician AAC Fanatic!

Oct 9, 2007
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I think it should be written:

$Au=\frac{1}{1+j \omega R1(C1+C2)}*Km*\frac{R3}{R2+R3}$

Why leave those superfluous factors in the expression?

Last edited: Jun 15, 2009
4. Jony130 AAC Fanatic!

Feb 17, 2009
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Well, in order to not give the direct answer.

5. zdzislavv Thread Starter Member

Jun 5, 2009
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I didn't expect that it is so simple - just a voltage divider on the right side of the circuit . I was looking for some materials about this topic (useful for exactly this kind of exercise), I found http://en.wikipedia.org/wiki/RC_circuit. I was also looking for it here http://users.ece.gatech.edu/~mleach/ but I couldn't find. Can you suggest me any good site with theoretical background concerning this topic, especially corner frequency (upper and lower) because I don't know how to proceed further with solving this exercise ?
And I don't know how you obtained your formula, Jony130. I tried (after reading your posts) to solve it in this way and I obtained directly The Electrician's solution.
http://images38.fotosik.pl/141/c954d5054975c6c0.jpg
And, by the way, what about Rin? Is there any resistance inside this amplifier which should be taken into account? What kind of amplifier is it?
Greetings!

6. The Electrician AAC Fanatic!

Oct 9, 2007
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If you examine the circuit, you will see that the amplifier is a simple buffer; there is no feedback around the amplifier, so R1 and C1,C2 just form a simple voltage divider.

All you need to know with a circuit that has no feedback, and where all the sub-circuits are in cascade (one following another), is that you can treat the sub-circuits individually and then multiply the transfer functions of all of them to get the final result. This circuit consists of the input voltage divider, the amplifier, and the output voltage divider.

If the amplifier were not ideal, then you would have take into account the loading effect, Rin, of the amplifier.

If you don't have a simple cascade of sub-circuits, then you may have to use a more general method of solving the circuit, such as the nodal method or the loop method.

What you need to do now is to calculate the overall voltage gain, Uv, at zero frequency (DC, in other words; set ω = 0), and then vary the frequency (ω) until you find the frequency at which the gain has decreased by a factor of SQRT(.5) = .707 = -3 dB.

What Jony130 did is to multiply the numerator and denominator by the conjugate of the denominator:

$Au=\frac{1*(1-j \omega R1(C1+C2))}{(1+j \omega R1(C1+C2))*(1-j \omega R1(C1+C2)}*Km*\frac{R3}{R2+R3}$

He then transformed the denominator product according to a factoring rule you should have learned when you first studied algebra:

$(a+b)*(a-b) = (a^2-b^2)$

If you multiply out the terms in the denominator, the j operator vanishes. This is called "rationalizing" the fraction; the final result is that the j operator is only present in the numerator.

This is a problem of yours, zdzislavv; you should be the person telling us what kind of amplifier it is, and what its input impedance is. How could we know those things? The diagram gives no indication of them. As far as I can see, the amplifier is an ideal amplifier with a voltage gain of 100.

Jun 5, 2009
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8. The Electrician AAC Fanatic!

Oct 9, 2007
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This is a single-pole lowpass filter (http://en.wikipedia.org/wiki/Lowpass), so it only has one corner frequency (ignoring any frequency limitations of the amplifier).

You might also look up highpass and bandpass filters on Wikipedia.